Let $G$ be a group with identity element $e$, let $a \in G$ have order $n$. Let $m$ be an integer then prove that $a^m =e$ iff $n$ divides $m$

Question

I have encountered this question in a book. I have no clue how to approach the question. How should I go about it? I was thinking of maybe writing m in terms of n, since we have m ≡ 0(mod n).

I tried to contradict it by taking some n which does not divide m. I am not sure if that is the correct procedure

How is e^k = e, given e can be any element of the group and not essentially 1, and k can be any integer. — Danish Shakeel, Nov 10 '19 at 15:12
Your title says $e$ is the identity element; by the way, you should put the question in the body, and the title should be a description of what the question is about — J. W. Tanner, Nov 10 '19 at 15:13
Could you show $n$ is not the order of $a$ if $am=e$ and $n\nmid m$? — J. W. Tanner, Nov 10 '19 at 15:16
Yes, indeed. e is the identity but identity of a group with some operation can be anything. Since the operation has not been mentioned, how can we say e^k = e? — Danish Shakeel, Nov 10 '19 at 15:16
if $e$ is the identity element then always $ek=e$; you could prove that by induction using the facts that $e=e$ (base case) and $e$ is inert — J. W. Tanner, Nov 10 '19 at 15:18
Do you mean $, m\cdot a = e\ $ or $, a^m = e,\ $ i.e. are you using additive or multiplicative notation? — Bill Dubuque, Nov 10 '19 at 15:42
Ok, you wrotye it additively in the title. I will change that and my answer. — Bill Dubuque, Nov 10 '19 at 15:44
I tried to contradict it by taking some n which does not divide m. I am not sure if that is the correct procedure. — Danish Shakeel, Nov 10 '19 at 15:45

Bill Dubuque · Answer 1 · 2019-11-10T15:46:08.870

1

Hint $\ H := \{m\in\Bbb Z\ :\ a^m = 1\}\,$ is closed under subtraction, so is a subgroup $\,\{0\}\subsetneq H\subseteq \Bbb Z.\,$ Therefore $\,H = n\Bbb Z\,$ where $\,n = $ least element $>0\,$ of $\,H.\,$ So $\,a^m = 1\!\iff\! m\in n\Bbb Z\!\iff\! n\mid m$

Remark $ $ Though it is easy to verify the set forms a subgroup by said subgroup test, it is more conceptual to do so by viewing $H$ as the kernel of the group hom: $\, m\mapsto a^m\,$ from $\,\Bbb Z\to G,\,$ presuming that viewpoint is already known.

edited Nov 10 '19 at 15:46

answered Nov 10 '19 at 15:21

Bill Dubuque

272,048

I write $1,$ vs. $,e,$ as the neutral element. – Bill Dubuque Nov 10 '19 at 15:48

J. W. Tanner · Accepted Answer · 2019-11-10T18:35:26.347

1

$\impliedby:\;$ If $n|m,$ then $m=nk,$ so $a^m=a^{(nk)}={(a^n)}^k=e^k=e$.

$\implies:\;$ If $a^m=e$ and $n\nmid m$, then $0<\gcd(n,m)<n$ but $a^{\gcd(n,m)}=e,$ (*)

contradicting the assumption that $a$ has order $n$.

Addendum (from my comments, as suggested by Bill Dubuque)

to justify the inequalities and equality on the line marked (*) above:

$0<\gcd(n,m)<n,$ because $n\ne0$ so $0<\gcd(n,m),$ and if $\gcd(n,m)=n$ then we'd have $n|m$.

By Bezout's identity, $\gcd(n,m)=nx+my$, so $a^{\gcd(n,m)}=a^{nx+my}=(a^n)^x(a^m)^y=e^xe^y=e$.

edited Nov 10 '19 at 18:35

answered Nov 10 '19 at 16:12

J. W. Tanner

60,406

At this level one should justify the two claimed (in)equalities after "then" in direction $\Longrightarrow$. We could also use $, m\bmod n,$ vs. $\gcd(m,n),$ to achieve the descent there (which is a special case of the proof that subgroups / ideal of $\Bbb Z$ are cyclic / principal, via the Euclidean algorithm, either in slow (subtractive) form or faster mod / gcd form). – Bill Dubuque Nov 10 '19 at 18:00
justification: $n\ne0$ so $0<\gcd(n,m)$, and if $\gcd(n,m)=n$ then we'd have $n|m$ – J. W. Tanner Nov 10 '19 at 18:03
The inequality is the easy part. You still need to justify the equality $,a^{\gcd(n,m)} = e,,$ which is usually the most difficult part for students (not only proving it, but understanding how to discover such a proof - which is clarified when one proceeds more conceptually as in my answer). – Bill Dubuque Nov 10 '19 at 18:07
by Bezout, $\gcd(n,m)=nx+my$ so $a^{\gcd(n,m)}=a^{nx+my}=(a^n)^x(a^m)^y=e^xe^y=e$ – J. W. Tanner Nov 10 '19 at 18:17
That's one way. Ideally the proofs should be in the answer (or linked) not in comments, since comments are ephemeral, not searched, etc. – Bill Dubuque Nov 10 '19 at 18:27

Let $G$ be a group with identity element $e$, let $a \in G$ have order $n$. Let $m$ be an integer then prove that $a^m =e$ iff $n$ divides $m$

2 Answers2